Question

Write a Lewis structure for each of the following polyatomic ions. Show all bonding valence electron pairs as lines and all nonbonding valence electron pairs as dots. For those ions that exhibit resonance, draw the various possible resonance forms. a. chlorate ion, \(\mathrm{ClO}_{3}^{-}\) b. peroxide ion, \(\mathrm{O}_{2}^{2-}\) c. acetate ion, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

Short answer

The Lewis structures for the given polyatomic ions are: a. Chlorate ion, \(\mathrm{ClO}_{3}^{-}\): \[ \mathrm{Cl} \: - \mathrm{O} \,|| \. \mathrm{O} \,|| \. \mathrm{O} \hspace{4pt} {:} \] b. Peroxide ion, \(\mathrm{O}_{2}^{2-}\): \[ \mathrm{O} \,|| \. \mathrm{O} \hspace{4pt} {:} \] c. Acetate ion, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\): \[ :{\mathrm{C}}(\mathrm{H}_{3}) \mathrm{C}(\mathrm{O}^{-}) \mathrm{O}|| \hspace{4pt} \longleftrightarrow \hspace{4pt} :{\mathrm{C}}(\mathrm{H}_{3}) \mathrm{C}||(\mathrm{O}) \mathrm{O}^{-} \]

Step-by-step solution

a. Chlorate ion, \(\mathrm{ClO}_{3}^{-}\).

1. Count valence electrons: Chlorine has 7 valence electrons, each oxygen atom has 6 valence electrons, and we have one extra electron due to the negative charge on the ion. Thus, we have a total of 7 + 3(6) + 1 = 26 valence electrons. 2. Place the central atom: Chlorine is the least electronegative atom, so it will be in the center. Connect it with three oxygen atoms using one electron pair (a single bond) for each connection. 3. Distribute remaining electrons: We have used 6 electrons for the single bonds, so we have 26 - 6 = 20 electrons left. Distribute them as lone pairs: each oxygen atom gets 3 lone pairs (6 electrons), which accounts for all of the remaining 20 electrons. 4. Check formal charges: Formal charge on chlorine = 7 (its valence electrons) - 3 (bonds) - 4 (lone electrons) = 0; formal charge on each oxygen = 6 - 2 (bonds) - 4 (lone electrons) = 0. The overall charge of the ion is -1, which is consistent with the given ion. 5. Since all of the formal charges are zero and no more resonance structures can be drawn, the Lewis structure of the chlorate ion, \(\mathrm{ClO}_{3}^{-}\), is complete. \( \mathrm{Cl} \: - \mathrm{O} \,|| \. \mathrm{O} \,|| \. \mathrm{O} \hspace{4pt} {:}\)

b. Peroxide ion, \(\mathrm{O}_{2}^{2-}\)

1. Count valence electrons: Each oxygen atom has 6 valence electrons, and we have two extra electrons due to the 2- charge on the ion. Thus, we have a total of 2(6) + 2 = 14 valence electrons. 2. Connect the two oxygen atoms with a single bond, using 2 electrons. 3. Distribute remaining electrons: We have used 2 electrons for the single bond, so we have 14 - 2 = 12 electrons left. Distribute them as lone pairs: each oxygen atom gets 3 lone pairs (6 electrons), which accounts for the remaining 12 electrons. 4. Check formal charges: Formal charge on each oxygen = 6 - 1 (bonds) - 6 (lone electrons) = -1. The overall charge of the ion is -2, which is consistent with the given ion. 5. The Lewis structure for the peroxide ion, \(\mathrm{O}_{2}^{2-}\), is as follows: \[ \mathrm{O} \,|| \. \mathrm{O} \hspace{4pt} {:} \]

c. Acetate ion, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

1. Count valence electrons: Each carbon atom has 4 valence electrons, each hydrogen atom has 1 valence electron, each oxygen atom has 6 valence electrons, and we have one extra electron due to the negative charge on the ion. Thus, we have 2(4) + 3(1) + 2(6) + 1 = 24 valence electrons. 2. Place the central atoms: a carbon atom is bonded to three hydrogen atoms and another carbon atom. Connect them using one electron pair (a single bond) each. The second carbon atom is bonded to two oxygen atoms. Connect them using one electron pair (a single bond) each. 3. Distribute remaining electrons: We have used 10 electrons for the single bonds, so we have 24 - 10 = 14 electrons left. Distribute them as lone pairs: each hydrogen atom gets 0 lone pairs, the first carbon atom gets 0 lone pairs, one of the oxygen atoms gets 3 lone pairs (6 electrons), and the other oxygen atom gets 2 lone pairs (4 electrons). This accounts for the remaining 14 electrons. 4. Check formal charges: The formal charge on C, C, H, and O with double bond are 0, but on O with single bond = 6 - 1 (bonds) - 6 (lone electrons) = -1, which is not optimal. 5. Adjust the structure by creating a resonance structure: Move some lone pairs from the singly bonded oxygen to create a double bond with carbon while making the double bond between carbon and the other oxygen into a single bond. This will create resonance structures with both oxygens having a formal charge of 0. The Lewis structures of the acetate ion, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\), are: \[ :{\mathrm{C}}(\mathrm{H}_{3}) \mathrm{C}(\mathrm{O}^{-}) \mathrm{O}|| \hspace{4pt} \longleftrightarrow \hspace{4pt} :{\mathrm{C}}(\mathrm{H}_{3}) \mathrm{C}||(\mathrm{O}) \mathrm{O}^{-} \]

Key concepts

Polyatomic Ions

Polyatomic ions are ions with more than one atom bonded together that carry a net charge. This charge is not neutral but is either positive or negative, determined by the difference in the number of protons and electrons. A common example is the chlorate ion

• Chlorate ion: This is composed of one chlorine atom and three oxygen atoms. It carries a net negative charge (\(\text{ClO}_3^−\)), resulting from an extra electron.
• Peroxide ion: Consists of two oxygen atoms bonded together, carrying a charge of minus two (\(\text{O}_2^{2-}\)).
• Acetate ion: Made of two carbon atoms, three hydrogen atoms, and two oxygen atoms with a net negative charge (\(\text{C}_2\text{H}_3\text{O}_2^−\)).

Understanding polyatomic ions is crucial for drawing Lewis structures because it affects the total number of valence electrons involved. A polyatomic ion can exhibit resonance structures if necessary, as seen in cases of ions like the chlorate and acetate ions, adding depth to understanding their chemical behavior. This detailed network of atoms bonded together with a resulting charge is fundamental in explaining how molecules like these ions interact chemically.

Valence Electrons

Valence electrons are the outermost electrons of an atom, crucial for chemical bonding. They are the electrons that form bonds with other atoms.

• Counting valence electrons helps determine how these electrons will distribute to achieve full outer shells and stability.
• Consider the chlorine atom in the chlorate ion (\(\text{ClO}_3^−\)): it has 7 valence electrons. Each oxygen contributes 6 valence electrons, and there's an extra one from the negative charge. This results in a total of 26 valence electrons available.
• For the peroxide ion (\(\text{O}_2^{2-}\)): each oxygen atom has 6 valence electrons, with 2 extra in total from the 2- charge, resulting in 14 valence electrons.
• In the acetate ion (\(\text{C}_2\text{H}_3\text{O}_2^{-}\)): the total comes from 2 carbon atoms each with 4 valence electrons, 3 hydrogen atoms each with 1, 2 oxygen atoms each with 6, plus one extra for the charge - totaling 24 valence electrons.

This calculation and distribution of electrons is fundamental to constructing Lewis structures, as it helps predict the bonding and non-bonding pairs of electrons.

Resonance Structures

Resonance structures are a way of representing the structure of a molecule or ion when two or more valid Lewis structures can be drawn. These structures are not distinct forms but rather a hybrid of potential configurations that depict the delocalization of electrons.In the case of molecules or ions exhibiting resonance:

• The acetate ion (\(\text{C}_2\text{H}_3\text{O}_2^−\)) has resonance. When drawing the resonance forms, electrons are shifted between oxygens, making possible different configurations where double bonds and lone pairs swap positions, thus each oxygen atom has a chance of being double-bonded to carbon.
• Similarly, chlorate ion (\(\text{ClO}_3^−\)) does not have traditional resonances in this simple structure exercise, but analyzing the idea helps understand the bonding capacity and stability of its structure.

The resonance concept prevents confusion about fixed single or double bond placements by spreading electrons over several bonds, giving an accurate representation of stable forms. It contributes essential insights into concepts like molecule stability, reactivity, and physical properties.